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Answer 1 · 2016-10-17T10:06:23

Let the distance between p and q be S m and the accelaration of the particle is #" "a" m/s^2#. Let the total time to cover distance S is T s.

The particle starts with initial velocity #u=0# and describes 2 m in 1st sec.

So applying kinematic equation
#s=ut+1/2at^2#
#2=0xx1+1/2xxaxx1^2#
#=>a=4m/s^2#

Again #S=0xxT+1/2xx4xxT^2#

#=>S=2T^2#

Distance covered in #(T-3)#s

#S_((T-3))=0xx(T-3)+1/2xx4(T-3)^2=2(T-3)^2#

Si the distance covered in last 3 sec

#S-S_((T-3))=11/16S=2T^2-2(T-3)^2#

#=>11/16xx2T^2=2T^2-2(T-3)^2#

#=>(T-3)^2=5/16T^2#

#=>(T-3)=sqrt5/4T#

#=>(4-sqrt5)/4T=3#

#=>T=12/(4-sqrt5)s=12/11(4+sqrt5)s#

So the dustance between p and q

#S=2T^2=12^2/11^2(4+sqrt5)^2m=46.28m#